100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached 4.2 TrustPilot
logo-home
Exam (elaborations)

Solutions Manual Fundamental Mechanics of Fluids 4th Edition by I. G. Currie

Rating
-
Sold
-
Pages
177
Grade
A+
Uploaded on
04-10-2025
Written in
2025/2026

This is a complete solutions manual for Fundamental Mechanics of Fluids 4th Edition by I. G. Currie. It provides detailed, step-by-step answers to all exercises and problems.

Institution
Fundamental Mechanics Of Fluids
Course
Fundamental Mechanics of Fluids











Whoops! We can’t load your doc right now. Try again or contact support.

Written for

Institution
Fundamental Mechanics of Fluids
Course
Fundamental Mechanics of Fluids

Document information

Uploaded on
October 4, 2025
Number of pages
177
Written in
2025/2026
Type
Exam (elaborations)
Contains
Questions & answers

Subjects

Content preview

Solutions Manual
Fundamental Mechanics of Fluids 4th Edition

By
I. G. Currie

( All Chapters Included - 100% Verified Solutions )




1

, BASIC CONSERVATION LAWS


Problem 1.1




Inflow through x = constant:  u y z

Outflow through x   x = constant:  u y z  (  u  y  z)  x 
x

Net inflow through x = constant surfaces:  ( u ) x y  z 
x

Net inflow through y = constant surfaces:  ( v ) x y  z 
y

Net inflow through z = constant surfaces:  ( w) x y  z 
z
But the rate at which the mass is accumulating inside the control volume is:

(   x  y  z)
t
Then the equation of mass conservation becomes:

    
 x  y  z    ( u )  (  v)  (  w)   x  y  z 
t  x y z 

Taking the limits as the quantities  x,  y and  z become vanishingly small, we get:


   
 ( u)  (  v)  (  w)  0
t x y z




Page 1-1

2

,BASIC CONSERVATION LAWS


Problem 1.2




Inflow through R = constant:  u R R   z

Outflow through R   R = constant:  u R R   z  (  u R R   z )  R 
R

Net inflow through R = constant surfaces:  (  R u R )  R   z 
R

Net inflow through  = constant surfaces:  (  u )  R   z 


Net inflow through z = constant surfaces:  (  u z ) R  R   z 
z
But the rate at which the mass is accumulating inside the control volume is:

(  R  R   z )
t
Then the equation of mass conservation becomes:

    
R  R   z    (  Ru R )  (  u  )  R (  u z )   R   z 
t  R  z 

Taking the limits as the quantities  R,  and  z become vanishingly small, we get:

 1  1  
 ( R u R )  (  u )  (  u z )  0
t R R R  z




Page 1-2

3

, BASIC CONSERVATION LAWS


Problem 1.3




Inflow through r = constant:  u r r 2 sin   
Outflow through r   r = constant:  u r r 2 sin   

 (  r 2u r sin    )  r 
r

Net inflow through r = constant surfaces: (  r 2u r )sin   r   
r

Net inflow through  = constant surfaces:  (  u sin  ) r  r   


Net inflow through  = constant surfaces:  (  u  ) r  r   

But the rate at which the mass is accumulating inside the control volume is:

(  r 2 sin   r   )
t
Then the equation of mass conservation becomes:

 2    
r sin   r      (  r 2u r ) sin   r (  u  sin  )  r (  u  )   r   
t  r   

Taking the limits as the quantities  r,  and  become vanishingly small, we get:


 1  1  1 
 2 (  r 2u r )  (  u  sin  )  (  u )  0
t r r r sin   r sin  




Page 1-3

4
$28.49
Get access to the full document:

100% satisfaction guarantee
Immediately available after payment
Both online and in PDF
No strings attached

Get to know the seller
Seller avatar
reckmila

Get to know the seller

Seller avatar
reckmila Massachusetts Institute Of Technology
View profile
Follow You need to be logged in order to follow users or courses
Sold
2
Member since
2 months
Number of followers
0
Documents
28
Last sold
1 week ago
Miss Fullmark

High-quality solutions manuals crafted to help you master every chapter and score full marks.

0.0

0 reviews

5
0
4
0
3
0
2
0
1
0

Recently viewed by you

Why students choose Stuvia

Created by fellow students, verified by reviews

Quality you can trust: written by students who passed their tests and reviewed by others who've used these notes.

Didn't get what you expected? Choose another document

No worries! You can instantly pick a different document that better fits what you're looking for.

Pay as you like, start learning right away

No subscription, no commitments. Pay the way you're used to via credit card and download your PDF document instantly.

Student with book image

“Bought, downloaded, and aced it. It really can be that simple.”

Alisha Student

Frequently asked questions